THE UNIVERSITY OF SYDNEY
Math2968 Algebra (Advanced)
Semester 2
Tutorial Solutions Week 12
2008
1.
(for general discussion) Recall that a poset
L
is a
lattice
if
x
∧
y
= inf
{
x,y
}
and
x
∨
y
= sup
{
x,y
}
exist for all
x,y
∈
L
. Discuss the existence or otherwise of each of the following:
inf
{
x
}
,
sup
{
x
}
,
inf
∅
,
sup
∅
,
inf
L,
sup
L
Solution
Both the following always exist for
x
∈
L
:
inf
{
x
}
= inf
{
x,x
}
=
x
∧
x
=
x
and
sup
{
x
}
= sup
{
x,x
}
=
x
∨
x
=
x
The largest and smallest elements of
L
may or may not exist and, when they do, equal
sup
L
= inf
∅
and
inf
L
= sup
∅
respectively.
2.
Let
L
be a lattice and choose
x,y
∈
L
such that
x
≤
y
. Verify that the interval
[
x,y
] =
{
z
∈
L

x
≤
z
≤
y
}
is a sublattice of
L
.
Solution
By reFexivity,
x
≤
x
≤
y
, so
x
∈
[
x,y
]
n
=
∅
. Let
a,b
∈
[
x,y
], so
x
≤
a
≤
y
and
x
≤
b
≤
y .
In particular,
y
is an upper bound for
a
and
b
. But
a
∨
b
is the least upper bound, so
a
∨
b
≤
y
. In
particular also,
a
∨
b
is an upper bound for
a
, so
x
≤
a
≤
a
∨
b
≤
y .
By transitivity,
x
≤
a
∨
b
≤
y
, so that
a
∨
b
∈
[
x,y
]. Dually (interchanging
≤
with
≥
) we get that
a
∧
b
∈
[
x,y
]. Hence [
x,y
] is closed under join and meet so is a sublattice.
3.
Verify the following associativity law for joins in a lattice:
x
∨
(
y
∨
z
) = (
x
∨
y
)
∨
z
Write down the corresponding associativity law for meets and explain how it follows by duality.
Solution
Put
α
=
x
∨
(
y
∨
z
) and
β
= (
x
∨
y
)
∨
z
. Then
α
is an upper bound for
x
and
y
∨
z
, which in turn is an
upper bound for
y
and
z
. By transitivity,
α
is an upper bound for
x
,
y
and
z
. In particular,
x
∨
y
≤
α
,
since
x
∨
y
is the least upper bound for